A solid cylinder of mass M = 1.31 kg and radius R = 8.9 cm pivots on a thin, fixed, frictionless bearing (see Figure). A string wrapped around the cylinder pulls downward with a force F = 2.354 N. Consider that instead of the force F, a block with mass m = 0.24 kg is attached to the cylinder with a massless string (see Figure). What is now the magnitude of the angular acceleration of the cylinder?
How far does mass m travel downward between t = 0.00 s and t = 0.20 s (Assuming motion begins at time t = 0.0 s)?
The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.0348 m in a time interval of 0.20 s. Find Icm of the new cylinder.
To find the magnitude of the angular acceleration of the cylinder, we can use Newton's second law for rotation:
τ = Iα
Where τ is the torque applied to the cylinder, I is the moment of inertia of the cylinder, and α is the angular acceleration.
In this case, the torque is equal to the force F multiplied by the radius R, and the moment of inertia of a solid cylinder rotating about its central axis is given by the formula:
I = (1/2)MR^2
Using these formulas, we can find the angular acceleration:
τ = FR = (2.354 N)(0.089 m) = 0.209 Nm
I = (1/2)(1.31 kg)(0.089 m)^2 = 0.00640 kgm^2
0.209 Nm = (0.00640 kgm^2)α
Simplifying the equation, we find:
α = 0.209 Nm / 0.00640 kgm^2 = 32.7 rad/s^2
Therefore, the magnitude of the angular acceleration of the cylinder is 32.7 rad/s^2.
To find how far mass m travels downward between t = 0.00 s and t = 0.20 s, we can use the kinematic equation for linear motion:
Δx = v0t + (1/2)at^2
Where Δx is the change in position (the distance traveled), v0 is the initial velocity (which is assumed to be zero in this case), a is the acceleration, and t is the time interval.
In this case, the acceleration is equal to the gravitational acceleration g, which is approximately 9.8 m/s^2.
Δx = (1/2)(9.8 m/s^2)(0.20 s)^2 = 0.196 m
Therefore, mass m travels 0.196 meters downward between t = 0.00 s and t = 0.20 s.
To find Icm of the new cylinder, we can use the equation for the average velocity:
v_avg = Δx / Δt
Where v_avg is the average velocity, Δx is the change in position (the distance traveled), and Δt is the time interval.
In this case, the average velocity is equal to the distance traveled divided by the time interval:
v_avg = 0.0348 m / 0.20 s = 0.174 m/s
We can use the formula for the average velocity of a uniformly accelerated motion:
v_avg = (v0 + v) / 2
Where v0 is the initial velocity (which is assumed to be zero in this case) and v is the final velocity.
Simplifying the equation, we find:
v = 2v_avg = 2(0.174 m/s) = 0.348 m/s
We can use the formula for linear velocity on a rotating cylinder:
v = ωR
Where v is the linear velocity, ω is the angular velocity, and R is the radius of the cylinder.
Rearranging the equation, we find:
ω = v / R = 0.348 m/s / 0.089 m = 3.91 rad/s
Finally, to find the moment of inertia of the new cylinder, we can use the formula for a solid cylinder rotating about its central axis:
I = (1/2)MR^2
Rearranging the equation, we find:
I = 2mR^2 = 2(0.24 kg)(0.089 m)^2 = 0.00305 kgm^2
Therefore, Icm of the new cylinder is 0.00305 kgm^2.